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I recently came across a math probability problem.

The problem says that is a party that has 6 attendees. Each attendee is asked to bring 1 food item from a list of 10 items. Assuming each attendee randomly chooses a food item to bring, what is the probability that at least 2 of the 6 cast members bring the same item?

I approached this by applying the logic that the P(2+ of 6 attendees bring in the same item) = 1 - P(everyone brings in different items).

Then, I decided in this problem, order does not matter; what matters is what the attendees bring in. Thus, I got the equation P(everyone brings in different items) = total possibilities where everyone brings in a different item/total 6 item combinations (with repetition). The first part is equal to 10C6 and the second part is equal to (10+6-1)!/6!9!. After some math, I eventually get 137/143.

This answer is wrong apparently.

The right answer is 1061/1250, solved by doing 10P6/10^6. I am confused because i feel like this is a problem where order doesn't matter, and thus should be a combination problem, yet the answer uses permutation formulas.

I still believe that my initial idea that "P(everyone brings in different items) = total possibilities where everyone brings in a different item/total 6 item combinations (with repetition)" is correct, but I don't know how to fit that into a "permutation oriented formula" (sorry, didn't know what to call it).

Can someone please tell me why combination is wrong (because it clearly is).

Whether my initial idea is correct. If so, an explanation on how you would use permutations with it would be appreciated. If not, then please present the correct answer and why it is correct.

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    $\begingroup$ Using nCr is definitely reliable when it is appropriate. The question is when it is appropriate. Bad title. $\endgroup$ Commented Apr 25, 2017 at 4:09

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The problem with your approach is that some combinations are more likely than others. The chance that you get items $1,2,3,4,5,6$ is $6!$ times higher than the chance you get six of item $1$. Your approach weights them the same. Try it by hand with three people bringing one of three items. The chance you get three different is $\frac 6{27}$ What do you get the chance to be?

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Finding probabilities by counting possibilities only works in that way when the possibilities are equally likely.

Your analysis boils down to "There are three possibilities for tossing two coins, therefore the chance of HH is 1/3".

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